# Question #90093

##### 1 Answer

#### Explanation:

Before moving on, it's worth mentioning that the *actual* value for the standard enthalpy of combustion of butane is

#DeltaH_"comb"^@ = - "2877.5 kJ mol"^(-1)#

https://en.wikipedia.org/wiki/Butane_(data_page)

I'll use the actual value in my calculations, but you can replace it with the value given to you if needed.

So, the first thing to do here is to convert the standard enthalpy of combustion of butane from *kilojoules per mole* to *kilojoules per gram*.

The **molar mass** of butane can be calculated by using

#M_ ("M C"_ 4"H"_ 10) = 4 xx M_ "M C" + 10 xx M_ "M H"#

This will get you

#M_ ("M C" _ 4"H"_ 10) = 4 xx "12 g mol"^(-1) + 10 xx "1 g mol"^(-1)#

#M_ ("M C" _ 4"H"_ 10) = "58 g mol"^(-1)#

Now, you know that the standard enthalpy of combustion of butane is equal to

#DeltaH_"comb"^@ = - "2877.5 kJ mol"^(-1)#

This tells you that when **mole** of butane undergoes combustion, **given off**, hence the *minus sign*.

Use the **molar mass** of the compound to convert this to *kilojoules per gram*

#-"2877.5 kJ"/(1 color(red)(cancel(color(black)("mole")))) * (1color(red)(cancel(color(black)("mole C"_4"H"_10))))/("58 g") = -"49.612 kJ g"^(-1)#

This tells you that when

You can thus say that the enthalpy change of combustion for your sample will be equal to

#14.5 color(red)(cancel(color(black)("g"))) * (-"49.612 kJ")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(-"719 kJ")))#

The answer is rounded to three **sig figs**.